\newproblem{lay:6_3_15}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.3.15}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $\mathbf{y}=(5,-9,5)$, $\mathbf{u}_1=(-3,-5,1)$ and $\mathbf{u}_2=(-3,2,1)$. Find the distance from $\mathbf{y}=(5,-9,5)$ to the plane in $\mathbb{R}^3$
	spanned by $\mathbf{u}_1$ and $\mathbf{u}_2$.
}{
   % Solution
	We project $\mathbf{y}$ onto $\mathrm{Span}\{\mathbf{u}_1,\mathbf{u}_2\}$
	\begin{center}
		$\begin{array}{rcl}
			\mathbf{y}_W&=&\frac{\mathbf{x}\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1+
			               \frac{\mathbf{x}\cdot\mathbf{u}_2}{\mathbf{u}_2\cdot\mathbf{u}_2}\mathbf{u}_2\\
									&=&\frac{35}{35}\begin{pmatrix}-3\\-5\\1\end{pmatrix}+
										 \frac{-28}{14}\begin{pmatrix}-3\\2\\1\end{pmatrix}
									 =\begin{pmatrix}3\\ -9\\ 1\end{pmatrix}\\
		\end{array}$
	\end{center}
	To find the vector perpendicular to $W$, we simply calculate
	\begin{center}
		$\begin{array}{rcl}
			\mathbf{y}_{W^\perp}&=&\mathbf{y}-\mathbf{y}_W=\begin{pmatrix}2\\ 0\\ 6\end{pmatrix}\\
		\end{array}$
	\end{center}
	The required distance is simply the norm of this vector that is $\sqrt{40}$.
}
\useproblem{lay:6_3_15}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
